inequalities

**jam2011** · January 31st 2011, 09:16 PM

Please check my work.thank you so much

**Prove It** · January 31st 2011, 09:27 PM

Quadratic Inequalities are always easiest to solve if you complete the square.

$\displaystyle x^2 - 2x - 3 \leq 0$

$\displaystyle x^2 - 2x + (-1)^2 - (-1)^2 - 3 \leq 0$

$\displaystyle (x - 1)^2 - 4 \leq 0$

$\displaystyle (x - 1)^2 \leq 4$

$\displaystyle \sqrt{(x-1)^2} \leq \sqrt{4}$

$\displaystyle |x-1| \leq 2$

$\displaystyle -2 \leq x - 1 \leq 2$

$\displaystyle -1 \leq x \leq 3$ .

**HallsofIvy** · February 1st 2011, 10:47 AM

Another good way to solve quadratic inequalities (and other complicated inequalities) is to solve the associated equation.

For example, to solve $\displaystyle x^2 - 2x - 3 \leq 0$ , first solve $\displaystyle x^2 - 2x - 3= 0$ , by factoring, $(x- 3)(x+ 1)= 0$ so x= 3 and x= -1. The point is this: a polynomial is a "continuous" function which means that it in order to go from positive to negative, it must pass through 0 and that can happen only at x= -1 and x= 3 which divide the real line into three intervals.

To decide which interval is ">0" and which "< 0" you can use either of two methods:
1) Choose a number in each interval and check the sign. x= -2 is less than -1 and $(-2)^2- 2(-2)- 3= 4+ 4- 3= 5> 0$ so the polynomial is greater than 0 for all x< -2. x= 0 is between -1 and 3 and $(0)^2- 2(0)- 3=-3< 0$ so the polynomial is less than 0 for all x between -1 and 3. Finally, x= 4 is larger than 3 and $(4)^2- 2(4)- 3= 5> 0$ so the polynomial is greater than 0 for all x greater than 3. Since the inequality say " $\le 0$ ", the solution set is $-1\le x\le 3$ .

2) Notice that x+ 1 is negative for x< -1 and positive for x> -1 and that x- 3 is negative for x< 3 and positive for x> 3. If x< -1, it is also less than 3 so both factors are negative and their product is positive. If -1< x< 3, one factor, x+1, is positive but x- 3 is still negative. The product of one positive and one negative factor is negative. Finally, if x> 3, both factors are positive and their product is positive. That gives the same solution set as before and (fortunately!) the same solution as Prove It.

**Archie Meade** · February 1st 2011, 12:38 PM

Originally Posted by jam2011

Please check my work.thank you so much

$x^2-2x-3\ \le\ 0$

$(x-3)(x+1)\le\ 0$

Your Case 1:

If $x-3 \ge\ 0$ , then $x+1=(x-3)+4$ must be $\ge\ 0$ since it is 4 greater than (x-3)

No luck there...

Case 2:

If $x-3\ \le\ 0$ , then $x\ \le\ 3$ and $x+1\ \ge\ 0$ only for $x\ \ge\ -1$

Hence $-1\ \le\ x\ \le\ 3$

so your answer is correct and the diagrams help illustrate the intersections.

Yet another way to proceed is....

$x^2-2x\ \le\ 3\Rightarrow\ x(x-2)\ \le\ 3$

The factors of 3 that differ by 2 are 1 and 3 or $-1$ and $-3$

$x(x-2)=3$ for $x=3$ and for $x=-1$

If $x<-1$ , then $x(x-2)>3$

If $x>3$ , then $x(x-2)>3$

The solution is $-1\ \le\ x\ \le\ 3$

**skeeter** · February 1st 2011, 02:03 PM

$x^2 - 2x - 3 \le 0$

$(x + 1)(x - 3) \le 0$

think about the graph of the quadratic ... a parabola that opens upward with two zeros, $x = -1$ and $x = 3$

the section of the graph below the x-axis (the less than or equal to section) lies between and includes the two zeros.

**jam2011** · February 1st 2011, 02:08 PM

nice job !!! thank you so much...i learn a lot.there are many ways to find and solve this problem. i owe you a lot...

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